DIVIDED POWER ALGEBRA
09PD
Contents
1. Introduction
2. Divided powers
3. Divided power rings
4. Extending divided powers
5. Divided power polynomial algebras
6. Tate resolutions
7. Application to complete intersections
8. Local complete intersection rings
9. Local complete intersection maps
10. Other chapters
References
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1. Introduction
09PE
In this chapter we talk about divided power algebras and what you can do with
them. A reference is the book [Ber74].
2. Divided powers
07GK
In this section we collect some results on divided power rings. We will use the
convention 0! = 1 (as empty products should give 1).
07GL
Definition 2.1. Let A be a ring. Let I be an ideal of A. A collection of maps
γn : I → I, n > 0 is called a divided power structure on I if for all n ≥ 0, m > 0,
x, y ∈ I, and a ∈ A we have
(1) γ1 (x) = x, we also set γ0 (x) = 1,
(2) γn (x)γm (x) = (n+m)!
n!m! γn+m (x),
(3) γn (ax) = an γP
n (x),
(4) γn (x + y) = i=0,...,n γi (x)γn−i (y),
(5) γn (γm (x)) =
(nm)!
n!(m!)n γnm (x).
(nm)!
Note that the rational numbers (n+m)!
n!m! and n!(m!)n occurring in the definition are
in fact integers; the first is the number of ways to choose n out of n + m and the
second counts the number of ways to divide a group of nm objects into n groups
of m. We make some remarks about the definition which show that γn (x) is a
replacement for xn /n! in I.
07GM
Lemma 2.2. Let A be a ring. Let I be an ideal of A.
This is a chapter of the Stacks Project, version 82fa846, compiled on Mar 04, 2016.
1
DIVIDED POWER ALGEBRA
2
(1) If γ is a divided power structure1 on I, then n!γn (x) = xn for n ≥ 1, x ∈ I.
Assume A is torsion free as a Z-module.
(2) A divided power structure on I, if it exists, is unique.
(3) If γn : I → I are maps then
γ is a divided power structure ⇔ n!γn (x) = xn ∀x ∈ I, n ≥ 1.
(4) The ideal I has a divided power structure if and only if there exists a set of
generators xi of I as an ideal such that for all n ≥ 1 we have xni ∈ (n!)I.
Proof. Proof of (1). If γ is a divided power structure, then condition (2) (applied
to 1 and n − 1 instead of n and m) implies that nγn (x) = γ1 (x)γn−1 (x). Hence by
induction and condition (1) we get n!γn (x) = xn .
Assume A is torsion free as a Z-module. Proof of (2). This is clear from (1).
Proof of (3). Assume that n!γn (x) = xn for all x ∈ I and n ≥ 1. Since A ⊂ A ⊗Z Q
it suffices to prove the axioms (1) – (5) of Definition 2.1 in case A is a Q-algebra.
In this case γn (x) = xn /n! and it is straightforward to verify (1) – (5); for example,
(4) corresponds to the binomial formula
X
n!
xi y n−i
(x + y)n =
i!(n
−
i)!
i=0,...,n
We encourage the reader to do the verifications to make sure that we have the
coefficients correct.
Proof of (4). Assume we have generators xi of I as an ideal such that xni ∈ (n!)I
for all n ≥ 1. We claim that for all x ∈ I we have xn ∈ (n!)I. If the claim holds
then we can set γn (x) = xn /n! which is a divided power structure by (3). To prove
the claim we note that it holds for x = axi . Hence we see that the claim holds
for a set of generators of I as an abelian group. By induction on the length of an
expression in terms of these, it suffices to prove the claim for x + y if it holds for x
and y. This follows immediately from the binomial theorem.
07GN
Example 2.3. Let p be a prime number. Let A be a ring such that every integer
n not divisible by p is invertible, i.e., A is a Z(p) -algebra. Then I = pA has a
canonical divided power structure. Namely, given x = pa ∈ I we set
pn n
γn (x) =
a
n!
The reader verifies immediately that pn /n! ∈ pZ(p) for n ≥ 1 (for instance, this
can be derived from the fact that the exponent of p in the prime factorization of
n! is bp/nc + p/n2 + p/n3 + . . .), so that the definition makes sense and gives
us a sequence of maps γn : I → I. It is a straightforward exercise to verify that
conditions (1) – (5) of Definition 2.1 are satisfied. Alternatively, it is clear that the
definition works for A0 = Z(p) and then the result follows from Lemma 4.2.
We notice that γn (0) = 0 for any ideal I of A and any divided power structure γ
on I. (This follows from axiom (3) in Definition 2.1, applied to a = 0.)
07GP
Lemma 2.4. Let A be a ring. Let I be an ideal of A. Let γn : I → I, n ≥ 1 be a
sequence of maps. Assume
1Here and in the following, γ stands short for a sequence of maps γ , γ , γ , . . . from I to I.
1 2 3
DIVIDED POWER ALGEBRA
3
(a) (1), (3), and (4) of Definition 2.1 hold for all x, y ∈ I, and
(b) properties (2) and (5) hold for x in some set of generators of I as an ideal.
Then γ is a divided power structure on I.
Proof. The numbers (1), (2), (3), (4), (5) in this proof refer to the conditions listed
in Definition 2.1. Applying (3) we see that if (2) and (5) hold for x then (2) and
(5) hold for ax for all a ∈ A. Hence we see (b) implies (2) and (5) hold for a set
of generators of I as an abelian group. Hence, by induction of the length of an
expression in terms of these it suffices to prove that, given x, y ∈ I such that (2)
and (5) hold for x and y, then (2) and (5) hold for x + y.
Proof of (2) for x + y. By (4) we have
X
γn (x + y)γm (x + y) =
i+j=n, k+l=m
Using (2) for x and y this equals
X (i + k)! (j + l)!
i!k!
Comparing this with the expansion
j!l!
γn+m (x + y) =
γi (x)γk (x)γj (y)γl (y)
γi+k (x)γj+l (y)
X
γa (x)γb (y)
we see that we have to prove that given a + b = n + m we have
X
(i + k)! (j + l)!
(n + m)!
=
.
i+k=a, j+l=b, i+j=n, k+l=m
i!k!
j!l!
n!m!
Instead of arguing this directly, we note that the result is true for the ideal I = (x, y)
in the polynomial ring Q[x, y] because γn (f ) = f n /n!, f ∈ I defines a divided power
structure on I. Hence the equality of rational numbers above is true.
Proof of (5) for x + y given that (1) – (4) hold and that (5) holds for x and y. We
will again reduce the proof to an
Pequality of rational numbers. Namely, using (4) we
can write γn (γm (x + y)) = γn ( γi (x)γj (y)). Using (4) we can write γn (γm (x + y))
as a sum of terms which are products of factors of the form γk (γi (x)γj (y)). If i > 0
then
γk (γi (x)γj (y)) = γj (y)k γk (γi (x))
(ki)!
γj (y)k γki (x)
k!(i!)k
(ki)! (kj)!
=
γki (x)γkj (y)
k!(i!)k (j!)k
=
using (3) in the first equality, (5) for x in the second, and (2) exactly k times in
the third. Using (5) for y we see the same equality holds when i = 0. Continuing
like this using all axioms but (5) we see that we can write
X
γn (γm (x + y)) =
cij γi (x)γj (y)
i+j=nm
for certain universal constants cij ∈ Z. Again the fact that the equality is valid
in the polynomial ring Q[x, y] implies that the coefficients cij are all equal to
(nm)!/n!(m!)n as desired.
07GQ
Lemma 2.5. Let A be a ring with two ideals I, J ⊂ A. Let γ be a divided power
structure on I and let δ be a divided power structure on J. Then
DIVIDED POWER ALGEBRA
4
(1) γ and δ agree on IJ,
(2) if γ and δ agree on I ∩ J then they are the restriction of a unique divided
power structure on I + J.
Proof. Let x ∈ I and y ∈ J. Then
γn (xy) = y n γn (x) = n!δn (y)γn (x) = δn (y)xn = δn (xy).
Hence γ and δ agree on a set of (additive) generators of IJ. By property (4) of
Definition 2.1 it follows that they agree on all of IJ.
Assume γ and δ agree on I ∩ J. Let z ∈ I + J. Write z = x + y with x ∈ I and
y ∈ J. Then we set
X
n (z) =
γi (x)δn−i (y)
for all n ≥ 1. To see that this is well defined, suppose that z = x0 + y 0 is another
representation with x0 ∈ I and y 0 ∈ J. Then w = x − x0 = y 0 − y ∈ I ∩ J. Hence
X
X
γi (x)δj (y) =
γi (x0 + w)δj (y)
i+j=n
i+j=n
X
=
γi0 (x0 )γl (w)δj (y)
i0 +l+j=n
X
=
γi0 (x0 )δl (w)δj (y)
i0 +l+j=n
X
=
γi0 (x0 )δj 0 (y + w)
i0 +j 0 =n
X
=
γi0 (x0 )δj 0 (y 0 )
0
0
i +j =n
as desired. Hence, we have defined maps n : I + J → I + J for all n ≥ 1; it is
easy to see that n |I = γn and n |J = δn . Next, we prove conditions (1) – (5) of
Definition 2.1 for the collection of maps n . Properties (1) and (3) are clear. To
see (4), suppose that z = x + y and z 0 = x0 + y 0 with x, x0 ∈ I and y, y 0 ∈ J and
compute
X
n (z + z 0 ) =
γa (x + x0 )δb (y + y 0 )
a+b=n
X
=
γi (x)γi0 (x0 )δj (y)δj 0 (y 0 )
i+i0 +j+j 0 =n
X
X
X
=
γi (x)δj (y)
γi0 (x0 )δj 0 (y 0 )
k=0,...,n
i+j=k
i0 +j 0 =n−k
X
=
k (z)n−k (z 0 )
k=0,...,n
as desired. Now we see that it suffices to prove (2) and (5) for elements of I or J,
see Lemma 2.4. This is clear because γ and δ are divided power structures.
The existence of a divided power structure on I + J whose restrictions to I and
J are γ and δ is thus proven; its uniqueness is rather clear.
07GR
Lemma 2.6. Let p be a prime number. Let A be a ring, let I ⊂ A be an ideal,
and let γ be a divided power structure on I. Assume p is nilpotent in A/I. Then I
is locally nilpotent if and only if p is nilpotent in A.
Proof. If pN = 0 in A, then for x ∈ I we have xpN = (pN )!γpN (x) = 0 because
(pN )! is divisible by pN . Conversely, assume I is locally nilpotent. We’ve also
assumed that p is nilpotent in A/I, hence pr ∈ I for some r, hence pr nilpotent,
hence p nilpotent.
DIVIDED POWER ALGEBRA
5
3. Divided power rings
07GT
There is a category of divided power rings. Here is the definition.
07GU
Definition 3.1. A divided power ring is a triple (A, I, γ) where A is a ring, I ⊂ A
is an ideal, and γ = (γn )n≥1 is a divided power structure on I. A homomorphism
of divided power rings ϕ : (A, I, γ) → (B, J, δ) is a ring homomorphism ϕ : A → B
such that ϕ(I) ⊂ J and such that δn (ϕ(x)) = ϕ(γn (x)) for all x ∈ I and n ≥ 1.
We sometimes say “let (B, J, δ) be a divided power algebra over (A, I, γ)” to indicate
that (B, J, δ) is a divided power ring which comes equipped with a homomorphism
of divided power rings (A, I, γ) → (B, J, δ).
07GV
Lemma 3.2. The category of divided power rings has all limits and they agree
with limits in the category of rings.
Proof. The empty limit is the zero ring (that’s weird but we need it).QThe product
Q
of a collection of divided power rings (At , It , γt ), t ∈ T is given by ( At , It , γ)
where γn ((xt )) = (γt,n (xt )). The equalizer of α, β : (A, I, γ) → (B, J, δ) is just
C = {a ∈ A | α(a) = β(a)} with ideal C ∩ I and induced divided powers. It follows
that all limits exist, see Categories, Lemma 14.10.
The following lemma illustrates a very general category theoretic phenomenon in
the case of divided power algebras.
07GW
Lemma 3.3. Let C be the category of divided power rings. Let F : C → Sets be a
functor. Assume that
(1) there exists a cardinal κ such that for every f ∈ F (A, I, γ) there exists
a morphism (A0 , I 0 , γ 0 ) → (A, I, γ) of C such that f is the image of f 0 ∈
F (A0 , I 0 , γ 0 ) and |A0 | ≤ κ, and
(2) F commutes with limits.
Then F is representable, i.e., there exists an object (B, J, δ) of C such that
F (A, I, γ) = HomC ((B, J, δ), (A, I, γ))
functorially in (A, I, γ).
Proof. This is a special case of Categories, Lemma 25.1.
07GX
Lemma 3.4. The category of divided power rings has all colimits.
Proof. The empty colimit is Z with divided power ideal (0). Let’s discuss general
colimits. Let C be a category and let c 7→ (Ac , Ic , γc ) be a diagram. Consider the
functor
F (B, J, δ) = limc∈C Hom((Ac , Ic , γc ), (B, J, δ))
Note that any f = (fc )c∈C ∈ F (B, J, δ) has the property that all the images fc (Ac )
generate a subring B 0 of B of bounded cardinality κ and that all the images fc (Ic )
generate a divided power sub ideal J 0 of B 0 . And we get a factorization of f as a f 0
in F (B 0 ) followed by the inclusion B 0 → B. Also, F commutes with limits. Hence
we may apply Lemma 3.3 to see that F is representable and we win.
DIVIDED POWER ALGEBRA
07GY
6
Remark 3.5. The forgetful functor (A, I, γ) 7→ A does not commute with colimits.
For example, let
/ (B 00 , J 00 , δ 00 )
(B, J, δ)
O
O
/ (B 0 , J 0 , δ 0 )
(A, I, γ)
be a pushout in the category of divided power rings. Then in general the map
B ⊗A B 0 → B 00 isn’t an isomorphism. (It is always surjective.) An explicit example
is given by (A, I, γ) = (Z, (0), ∅), (B, J, δ) = (Z/4Z, 2Z/4Z, δ), and (B 0 , J 0 , δ 0 ) =
(Z/4Z, 2Z/4Z, δ 0 ) where δ2 (2) = 2 and δ20 (2) = 0 and all higher divided powers
equal to zero. Then (B 00 , J 00 , δ 00 ) = (F2 , (0), ∅) which doesn’t agree with the tensor
product. However, note that it is always true that
B 00 /J 00 = B/J ⊗A/I B 0 /J 0
as can be seen from the universal property of the pushout by considering maps into
divided power algebras of the form (C, (0), ∅).
4. Extending divided powers
07GZ
Here is the definition.
07H0
Definition 4.1. Given a divided power ring (A, I, γ) and a ring map A → B we
say γ extends to B if there exists a divided power structure γ̄ on IB such that
(A, I, γ) → (B, IB, γ̄) is a homomorphism of divided power rings.
07H1
Lemma 4.2. Let (A, I, γ) be a divided power ring. Let A → B be a ring map.
If γ extends to B then it extends uniquely. Assume (at least) one of the following
conditions holds
(1) IB = 0,
(2) I is principal, or
(3) A → B is flat.
Then γ extends to B.
Pt
Proof. Any element of IB can be written as a finite sum i=1 bi xi with bi ∈ B
and xi ∈ I. If γ extends to γ̄ on IB then γ̄n (xi ) = γn (xi ). Thus, conditions (3)
and (4) in Definition 2.1 imply that
Xt
X
Yt
γ̄n (
bi xi ) =
bni i γni (xi )
i=1
n1 +...+nt =n
i=1
Thus we see that γ̄ is unique if it exists.
If IB = 0 then setting γ̄n (0) = 0 works. If I = (x) then we define γ̄n (bx) = bn γn (x).
This is well defined: if b0 x = bx, i.e., (b − b0 )x = 0 then
bn γn (x) − (b0 )n γn (x) = (bn − (b0 )n )γn (x)
= (bn−1 + . . . + (b0 )n−1 )(b − b0 )γn (x) = 0
because γn (x) is divisible by x (since γn (I) ⊆ I) and hence annihilated by b − b0 .
Next, we prove conditions (1) – (5) of Definition 2.1. Parts (1), (2), (3), (5) are
DIVIDED POWER ALGEBRA
7
obvious from the construction. For (4) suppose that y, z ∈ IB, say y = bx and
z = cx. Then y + z = (b + c)x hence
γ̄n (y + z) = (b + c)n γn (x)
X
n!
=
bi cn−i γn (x)
i!(n − i)!
X
=
bi cn−i γi (x)γn−i (x)
X
=
γ̄i (y)γ̄n−i (z)
as desired.
Assume A → B is flat. Suppose that b1 , . . . , br ∈ B and x1 , . . . , xr ∈ I. Then
X
X
γ̄n (
bi x i ) =
be11 . . . berr γe1 (x1 ) . . . γer (xr )
where the sum is over e1 + . . . + er = n ifPγ̄n exists. Next suppose
P that we have
c1 , . . . , cs ∈ B and aij ∈ A such that bi =
aij cj . Setting yj =
aij xi we claim
that
X
X
be11 . . . berr γe1 (x1 ) . . . γer (xr ) =
cd11 . . . cds s γd1 (y1 ) . . . γds (ys )
in B where on the right hand side we are summing over d1 + . . . + ds = n. Namely,
using the axioms of a divided power structure we can expand both sides into a sum
with coefficients in Z[aij ] of terms of the form cd11 . . . cds s γe1 (x1 ) . . . γer (xr ). To see
that the coefficients agree we note that the result is true in Q[x1 , . . . , xr , c1 , . . . , cs , aij ]
with γ the unique divided power structure on (x1 , . . . , xr ). By Lazard’s theorem
(Algebra, Theorem 80.4) we can write B as a directedPcolimit of finitePfree Amodules. In particular, if z ∈ IB is written as z =
xi bi and z P
=
x0i0 b0i0 ,
0
then we can find c1 , . . . , cs ∈ B and aij , ai0 j ∈ A such that bi =
aij cj and
P 0
P
P 0 0
b0i0 =
ai0 j cj such that yj =
xi aij =
xi0 ai0 j holds2. Hence the procedure
above gives a well defined map γ̄n on IB. By construction γ̄ satisfies conditions
(1), (3), and (4). Moreover, for x ∈ I we have γ̄n (x) = γn (x). Hence it follows from
Lemma 2.4 that γ̄ is a divided power structure on IB.
07H2
Lemma 4.3. Let (A, I, γ) be a divided power ring.
(1) If ϕ : (A, I, γ) → (B, J, δ) is a homomorphism of divided power rings, then
Ker(ϕ) ∩ I is preserved by γn for all n ≥ 1.
(2) Let a ⊂ A be an ideal and set I 0 = I ∩ a. The following are equivalent
(a) I 0 is preserved by γn for all n > 0,
(b) γ extends to A/a, and
(c) there exist a set of generators xi of I 0 as an ideal such that γn (xi ) ∈ I 0
for all n > 0.
Proof. Proof of (1). This is clear. Assume (2)(a). Define γ̄n (x mod I 0 ) = γn (x) mod
I 0 for x ∈ I. This is well defined since γn (x + y) = γn (x) mod I 0 for y ∈ I 0 by Definition 2.1 (4) and the fact that γj (y) ∈ I 0 by assumption. It is clear that γ̄ is
a divided power structure as γ is one. Hence (2)(b) holds. Also, (2)(b) implies
(2)(a) by part (1). It is clear that (2)(a) implies (2)(c). Assume (2)(c). Note that
2This can also be proven without recourse to Algebra, Theorem 80.4. Indeed, if z = P x b
i i
P 0 0
P
P 0 0
and z =
xi0 bi0 , then
xi bi −
xi0 bi0 = 0 is a relation in the A-module B. Thus, Algebra,
Lemma 38.11 (applied to the xi and x0i0 taking the place of the fi , and the bi and b0i0 taking the
role of the xi ) yields the existence of the c1 , . . . , cs ∈ B and aij , a0i0 j ∈ A as required.
DIVIDED POWER ALGEBRA
8
γn (x) = an γn (xi ) ∈ I 0 for x = axi . Hence we see that γn (x) ∈ I 0 for a set of
generators of I 0 as an abelian group. By induction on the length of an expression
in terms of these, it suffices to prove ∀n : γn (x + y) ∈ I 0 if ∀n : γn (x), γn (y) ∈ I 0 .
This follows immediately from the fourth axiom of a divided power structure. 07H3
Lemma 4.4. Let (A, I, γ) be a divided power ring. Let E ⊂ I be a subset. Then
the smallest ideal J ⊂ I preserved by γ and containing all f ∈ E is the ideal J
generated by γn (f ), n ≥ 1, f ∈ E.
Proof. Follows immediately from Lemma 4.3.
07KD
Lemma 4.5. Let (A, I, γ) be a divided power ring. Let p be a prime. If p is
nilpotent in A/I, then
(1) the p-adic completion A∧ = lime A/pe A surjects onto A/I,
(2) the kernel of this map is the p-adic completion I ∧ of I, and
(3) each γn is continuous for the p-adic topology and extends to γn∧ : I ∧ → I ∧
defining a divided power structure on I ∧ .
If moreover A is a Z(p) -algebra, then
(4) for e large enough the ideal pe A ⊂ I is preserved by the divided power
structure γ and
(A∧ , I ∧ , γ ∧ ) = lime (A/pe A, I/pe A, γ̄)
in the category of divided power rings.
Proof. Let t ≥ 1 be an integer such that pt A/I = 0, i.e., pt A ⊂ I. The map
A∧ → A/I is the composition A∧ → A/pt A → A/I which is surjective (for example
by Algebra, Lemma 95.1). As pe I ⊂ pe A ∩ I ⊂ pe−t I for e ≥ t we see that the
kernel of the composition A∧ → A/I is the p-adic completion of I. The map γn is
continuous because
X
γn (x + pe y) =
pje γi (x)γj (y) = γn (x) mod pe I
i+j=n
by the axioms of a divided power structure. It is clear that the axioms for divided
power structures are inherited by the maps γn∧ from the maps γn . Finally, to see
the last statement say e > t. Then pe A ⊂ I and γ1 (pe A) ⊂ pe A and for n > 1 we
have
pn n(e−1) n
γn (pe a) = pn γn (pe−1 a) =
p
a ∈ pe A
n!
as pn /n! ∈ Z(p) and as n ≥ 2 and e ≥ 2 so n(e − 1) ≥ e. This proves that γ extends
to A/pe A, see Lemma 4.3. The statement on limits is clear from the construction
of limits in the proof of Lemma 3.2.
5. Divided power polynomial algebras
07H4
A very useful example is the divided power polynomial algebra. Let A be a ring.
Let t ≥ 1. We will denote Ahx1 , . . . , xt i the following A-algebra: As an A-module
we set
M
[n ]
[n ]
Ahx1 , . . . , xt i =
Ax1 1 . . . xt t
n1 ,...,nt ≥0
with multiplication given by
[n] [m]
xi xi
=
(n + m)! [n+m]
xi
.
n!m!
DIVIDED POWER ALGEBRA
[1]
[0]
9
[0]
We also set xi = xi . Note that 1 = x1 . . . xt . There is a similar construction which gives the divided power polynomial algebra in infinitely many variables.
[n]
There is an canonical A-algebra map Ahx1 , . . . , xt i → A sending xi to zero for
n > 0. The kernel of this map is denoted Ahx1 , . . . , xt i+ .
07H5
Lemma 5.1. Let (A, I, γ) be a divided power ring. There exists a unique divided
power structure δ on
J = IAhx1 , . . . , xt i + Ahx1 , . . . , xt i+
such that
[n]
(1) δn (xi ) = xi , and
(2) (A, I, γ) → (Ahx1 , . . . , xt i, J, δ) is a homomorphism of divided power rings.
Moreover, (Ahx1 , . . . , xt i, J, δ) has the following universal property: A homomorphism of divided power rings ϕ : (Ahxi, J, δ) → (C, K, ) is the same thing as a
homomorphism of divided power rings A → C and elements k1 , . . . , kt ∈ K.
Proof. We will prove the lemma in case of a divided power polynomial algebra in
one variable. The result for the general case can be argued in exactly the same way,
or by noting that Ahx1 , . . . , xt i is isomorphic to the ring obtained by adjoining the
divided power variables x1 , . . . , xt one by one.
Let Ahxi+ be the ideal generated by x, x[2] , x[3] , . . .. Note that J = IAhxi + Ahxi+
and that
IAhxi ∩ Ahxi+ = IAhxi · Ahxi+
Hence by Lemma 2.5 it suffices to show that there exist divided power structures on
the ideals IAhxi and Ahxi+ . The existence of the first follows from Lemma 4.2 as
A → Ahxi is flat. For the second, note that if A is torsion free, then we can apply
Lemma 2.2 (4) to see that δ exists. Namely, choosing as generators the elements
[nm]
x[m] we see that (x[m] )n = (nm)!
and n! divides the integer (nm)!
(m!)n x
(m!)n . In general
write A = R/a for some torsion
free
ring
R
(e.g.,
a
polynomial
ring
over Z). The
L [m]
kernel of Rhxi → Ahxi is
ax . Applying criterion (2)(c) of Lemma 4.3 we see
that the divided power structure on Rhxi+ extends to Ahxi as desired.
Proof of the universal property. Given a homomorphism ϕ : A → C of divided
power rings and k1 , . . . , kt ∈ K we consider
Ahx1 , . . . , xt i → C,
[n1 ]
x1
[nt ]
. . . xt
7−→ n1 (k1 ) . . . nt (kt )
using ϕ on coefficients. The only thing to check is that this is an A-algebra homomorphism (details omitted). The inverse construction is clear.
07H6
Remark 5.2. Let (A, I, γ) be a divided power ring. There is a variant of Lemma
5.1 for infinitely many variables. First note that if s < t then there is a canonical
map
Ahx1 , . . . , xs i → Ahx1 , . . . , xt i
Hence if W is any set, then we set
Ahxw , w ∈ W i = colimE⊂W Ahxe , e ∈ Ei
(colimit over E finite subset of W ) with transition maps as above. By the definition of a colimit we see that the universal mapping property of Ahxw , w ∈ W i is
completely analogous to the mapping property stated in Lemma 5.1.
DIVIDED POWER ALGEBRA
10
The following lemma can be found in [BO83].
07GS
Lemma 5.3. Let p be a prime number. Let A be a ring such that every integer
n not divisible by p is invertible, i.e., A is a Z(p) -algebra. Let I ⊂ A be an ideal.
Two divided power structures γ, γ 0 on I are equal if and only if γp = γp0 . Moreover,
given a map δ : I → I such that
(1) p!δ(x) = xp for all x ∈ I,
(2) δ(ax) = ap δ(x) forP
all a ∈ A, x ∈ I, and
1 i j
(3) δ(x + y) = δ(x) + i+j=p,i,j≥1 i!j!
x y + δ(y) for all x, y ∈ I,
then there exists a unique divided power structure γ on I such that γp = δ.
Proof. If n is not divisible by p, then γn (x) = cxγn−1 (x) where c is a unit in Z(p) .
Moreover,
γpm (x) = cγm (γp (x))
where c is a unit in Z(p) . Thus the first assertion is clear. For the second assertion,
we can, working backwards, use these equalities to define all γn . More precisely, if
n = a0 + a1 p + . . . + ae pe with ai ∈ {0, . . . , p − 1} then we set
γn (x) = cn xa0 δ(x)a1 . . . δ e (x)ae
for cn ∈ Z(p) defined by
cn = (p!)a1 +a2 (1+p)+...+ae (1+...+p
e−1
)
/n!.
Now we have to show the axioms (1) – (5) of a divided power structure, see Definition 2.1. We observe that (1) and (3) are immediate. Verification of (2) and (5) is
by a direct calculation which we omit. Let x, y ∈ I. We claim there is a ring map
ϕ : Z(p) hu, vi −→ A
which maps u[n] to γn (x) and v [n] to γn (y). By construction of Z(p) hu, vi this means
we have to check that
γn (x)γm (x) =
(n + m)!
γn+m (x)
n!m!
in A and similarly for y. This is true because (2) holds for γ. Let denote the
divided power structure on the ideal Z(p) hu, vi+ of Z(p) hu, vi. Next, we claim
that ϕ(n (f )) = γn (ϕ(f )) for f ∈ Z(p) hu, vi+ and all n. This is clear for n =
0, 1, . . . , p − 1. For n = p it suffices to prove it for a set of generators of the ideal
Z(p) hu, vi+ because both p and γp = δ satisfy properties (1) and (3) of the lemma.
(pn)!
Hence it suffices to prove that γp (γn (x)) = p!(n!)
p γpn (x) and similarly for y, which
follows as (5) holds for γ. Now, if n = a0 + a1 p + . . . + ae pe is an arbitrary integer
written in p-adic expansion as above, then
n (f ) = cn f a0 γp (f )a1 . . . γpe (f )ae
because is a divided power structure. Hence we see that ϕ(n (f )) = γn (ϕ(f ))
holds for all n. Applying this for f = u + v we see that axiom (4) for γ follows from
the fact that is a divided power structure.
DIVIDED POWER ALGEBRA
11
6. Tate resolutions
09PF
In this section we briefly discuss the resolutions constructed in [Tat57] which combine divided power structures with differential graded algebras. In this section
we will use homological notation for differential graded algebras. Our differential
graded algebras will sit in nonnegative homological degrees. Thus our differential
graded algebras (A, d) will be given as chain complexes
. . . → A2 → A1 → A0 → 0 → . . .
endowed with a multiplication.
09PG
Let
L R be a ring. In this section we will often consider graded R-algebras A =
A whose components are zero in negative degrees. We will set A+ =
Ld≥0 d
L
L
d>0 Ad . We will write Aeven =
d≥0 A2d and Aodd =
d≥0 A2d+1 . Recall
deg(x) deg(y)
that A is graded commutative if xy = (−1)
yx for homogeneous elements
x, y. Recall that A is strictly graded commutative if in addition x2 = 0 for homogeneous elements x of odd degree. Finally, to understand the following definition,
keep in mind that γn (x) = xn /n! if A is a Q-algebra.
L
Definition 6.1. Let R be a ring. Let A = d≥0 Ad be a graded R-algebra which is
strictly graded commutative. A collection of maps γn : Aeven,+ → Aeven,+ defined
for all n > 0 is called a divided power structure on A if we have
(1) γn (x) ∈ A2nd if x ∈ A2d ,
(2) γ1 (x) = x for any x, we also set γ0 (x) = 1,
(3) γn (x)γm (x) = (n+m)!
n!m! γn+m (x),
(4) γn (xy) = xn γn (y) for all x ∈ Aeven and y ∈ Aeven,+ ,
(5) γn (xy) = 0 if x, y ∈ Aodd homogeneous
P and n > 1
(6) if x, y ∈ Aeven,+ then γn (x + y) = i=0,...,n γi (x)γn−i (y),
(7) γn (γm (x)) =
(nm)!
n!(m!)n γnm (x)
for x ∈ Aeven,+ .
Observe that conditions (2), (3), (4), (6), and (7) imply that γ is a “usual” divided
power structure on the ideal Aeven,+ of the (commutative) ring Aeven , see Sections
2, 3, 4, and 5. In particular, we have n!γn (x) = xn for all x ∈ Aeven,+ . Condition
(1) states that γ is compatible with grading and condition (5) tells us γn for n > 1
vanishes on products of homogeneous elements of odd degree. But note that it may
happen that
γ2 (z1 z2 + z3 z4 ) = z1 z2 z3 z4
is nonzero if z1 , z2 , z3 , z4 are homogeneous elements of odd degree.
09PH
Example 6.2 (Adjoining odd variable). Let R be a ring. Let (A, γ) be a strictly
graded commutative graded R-algebra endowed with a divided power structure as
in the definition above. Let d > 0 be an odd integer. In this setting we can adjoin
a variable T of degree d to A. Namely, set
AhT i = A ⊕ AT
with grading given by AhT im = Am ⊕ Am−d T . We claim there is a unique divided
power structure on AhT i compatible with the given divided power structure on A.
Namely, we set
γn (x + yT ) = γn (x) + γn−1 (x)yT
for x ∈ Aeven,+ and y ∈ Aodd .
DIVIDED POWER ALGEBRA
09PI
12
Example 6.3 (Adjoining even variable). Let R be a ring. Let (A, γ) be a strictly
graded commutative graded R-algebra endowed with a divided power structure as
in the definition above. Let d > 0 be an even integer. In this setting we can adjoin
a variable T of degree d to A. Namely, set
AhT i = A ⊕ AT ⊕ AT (2) ⊕ AT (3) ⊕ . . .
with multiplication given by
T (n) T (m) =
(n + m)! (n+m)
T
n!m!
and with grading given by
AhT im = Am ⊕ Am−d T ⊕ Am−2d T (2) ⊕ . . .
We claim there is a unique divided power structure on AhT i compatible with the
given divided power structure on A such that γn (T (i) ) = T (ni) . To define the
divided power structure we first set
X
XY
ei (iei )
γn
xi T (i) =
P xi T
i>0
n=
ei
if xi is in Aeven . If x0 ∈ Aeven,+ then we take
X
X
X
γn
xi T (i) =
γa (x0 )γb
i≥0
a+b=n
i>0
xi T (i)
where γb is as defined above.
09PJ
At this point we tie in the definition of divided power structures with differentials.
To understand the definition note that d(xn /n!) = d(x)xn−1 /(n − 1)! if A is a
Q-algebra and x ∈ Aeven,+ .
L
Definition 6.4. Let R be a ring. Let A =
d≥0 Ad be a differential graded
R-algebra which is strictly graded commutative. A divided power structure γ on
A is compatible with the differential graded structure if d(γn (x)) = d(x)γn−1 (x) for
all x ∈ Aeven,+ .
Warning: Let (A, d, γ) be as in Definition 6.4. It may not be true that γn (x)
is a boundary, if x is a boundary. Thus γ in general does not induce a divided
power structure on the homology algebra H(A). In some papers the authors put
an additional compatibility condition in order to insure this is the case, but we elect
not to do so.
09PK
Lemma 6.5. Let (A, d, γ) and (B, d, γ) be as in Definition 6.4. Let f : A → B
be a map of differential graded algebras compatible with divided power structures.
Assume
(1) Hk (A) = 0 for k > 0, and
(2) f is surjective.
Then γ induces a divided power structure on the graded R-algebra H(B).
Proof. Suppose that x and x0 are homogeneous of the same degree 2d and define
the same cohomology class in H(B). Say x0 − x = d(w). Choose a lift y ∈ A2d of
x and a lift z ∈ A2d+1 of w. Then y 0 = y + d(z) is a lift of x0 . Hence
X
X
γn (y 0 ) =
γi (y)γn−i (d(z)) = γn (y) +
γi (y)γn−i (d(z))
i<n
DIVIDED POWER ALGEBRA
13
Since A is acyclic in positive degrees and since d(γj (d(z))) = 0 for all j we can
write this as
X
γn (y 0 ) = γn (y) +
γi (y)d(zi )
i<n
for some zi in A. Moreover, for 0 < i < n we have
d(γi (y)zi ) = d(γi (y))zi + γi (y)d(zi ) = d(y)γi−1 (y)zi + γi (y)d(zi )
and the first term maps to zero in B as d(y) maps to zero in B. Hence γn (x0 )
and γn (x) map to the same element of H(B). Thus we obtain a well defined map
γn : H2d (B) → H2nd (B) for all d > 0 and n > 0. We omit the verification that this
defines a divided power structure on H(B).
09PL
Lemma 6.6. Let (A, d, γ) be as in Definition 6.4. Let R → R0 be a ring map.
Then d and γ induce similar structures on A0 = A ⊗R R0 such that (A0 , d, γ) is as
in Definition 6.4.
Proof. Observe that A0even = Aeven ⊗R R0 and A0even,+ = Aeven,+ ⊗R R0 . Hence
we are trying to show that the divided powers γ extend to A0even (terminology as in
Definition 4.1). Once we have shown γ extends it follows easily that this extension
has all the desired properties.
Choose a polynomial R-algebra P and a surjection of R-algebras P → R0 . The ring
map Aeven → Aeven ⊗R P is flat, hence the divided powers γ extend to Aeven ⊗R P
uniquely by Lemma 4.2. Let J = Ker(P → R0 ). To show that γ extends to A ⊗R R0
it suffices to show that I 0 = Ker(Aeven,+ ⊗R P → Aeven,+ ⊗R R0 ) is generated by
elements z such that γn (z) ∈ I 0 for all n > 0. This is clear as I 0 is generated by
elements of the form x ⊗ f with x ∈ Aeven,+ and f ∈ Ker(P → R0 ).
09PM
Lemma 6.7. Let (A, d, γ) be as in Definition 6.4. Let d ≥ 1 be an integer.
Let AhT i be the graded divided power polynomial algebra on T with deg(T ) = d
constructed in Example 6.2 or 6.3. Let f ∈ Ad−1 be an element with d(f ) = 0.
There exists a unique differential d on AhT i such that d(T ) = f and such that d is
compatible with the divided power structure on AhT i.
Proof. This is proved by a direct computation which is omitted.
Here is the construction of Tate.
09PN
Lemma 6.8. Let R be a Noetherian ring. Let R → S be of finite type. There
exists a factorization
R→A→S
with the following properties
(1)
(2)
(3)
(4)
(A, d, γ) is as in Definition 6.4,
A → S is a quasi-isomorphism (if we endow S with the zero differential),
A0 = R[x1 , . . . , xn ] → S is any surjection of a polynomial ring onto S, and
A is a graded divided power polynomial algebra over R with finitely many
variables in each degree.
The last condition means that A is constructed out of A0 by successively adjoining
variables T of degree > 0 as in Examples 6.2 and 6.3.
DIVIDED POWER ALGEBRA
14
Proof. Start of the construction. Let A(0) = R[x1 , . . . , xn ] be a (usual) polynomial
ring and let A(0) → S be a surjection. As grading we take A(0)0 = A(0) and
A(0)d = 0 for d 6= 0. Thus d = 0 and γn , n > 0 is zero as well.
Choose generators f1 , . . . , fm ∈ R[x1 , . . . , xm ] for the kernel of the given map
A(0) = R[x1 , . . . , xm ] → S. We apply Examples 6.2 m times to get
A(1) = A(0)hT1 , . . . , Tm i
with deg(Ti ) = 1 as a graded divided power polynomial algebra. We set d(Ti ) = fi .
Since A(1) is a divided power polynomial algebra over A(0) and since d(fi ) = 0
this extends uniquely to a differential on A(1) by Lemma 6.7.
Induction hypothesis: Assume we are given factorizations
R → A(0) → A(1) → . . . → A(m) → S
where A(0) and A(1) are as above and each R → A(m0 ) → S for 2 ≤ m0 ≤ m
satisfies properties (1) and (4) of the statement of the lemma and (2) replaced by
the condition that Hi (A(m0 )) → Hi (S) is an isomorphism for m0 > i ≥ 0. The
base case is m = 1.
Induction step. Assume we have R → A(m) → S as in the induction hypothesis.
Consider the group Hm (A(m)). This is a module over H0 (A(m)) = S. In fact, it is
a subquotient of A(m)m which is a finite type module over A(m)0 = R[x1 , . . . , xn ].
Thus we can pick finitely many elements
e1 , . . . , et ∈ Ker(d : A(m)m → A(m)m−1 )
which map to generators of this module. Applying Example 6.3 or 6.2 t times we
get
A(m + 1) = A(m)hT1 , . . . , Tt i
with deg(Ti ) = m + 1 as a graded divided power algebra. We set d(Ti ) = ei . Since
A(1) is a divided power polynomial algebra over A(0) and since d(ei ) = 0 this
extends uniquely to a differential on A(m + 1) compatible with the divided power
structure. Since we’ve added only material in degree m + 1 and higher we see that
Hi (A(m + 1)) = Hi (A(m)) for i < m. Moreover, it is clear that Hm (A(m + 1)) = 0
by construction.
To finish the proof we observe that we have shown there exists a sequence of maps
R → A(0) → A(1) → . . . → A(m) → A(m + 1) → . . . → S
and to finish the proof we set A = colim A(m).
0BZ9
Lemma 6.9. Let R → S be a pseudo-coherent ring map (More on Algebra, Definition 67.1). Then Lemma 6.8 holds.
Proof. This is proved in exactly the same way as Lemma 6.8. The only additional
twist is that, given A(m) → S we have to show that Hm = Hm (A(m)) is a finite
R[x1 , . . . , xm ]-module (so that in the next step we need only add finitely many
variables). Consider the complex
. . . → A(m)m−1 → A(m)m → A(m)m−1 → . . . → A(m)0 → S → 0
Since S is a pseudo-coherent R[x1 , . . . , xn -module and since A(m)i is a finite free
R[x1 , . . . , xn ]-module we conclude that this is a pseudo-coherent complex, see More
DIVIDED POWER ALGEBRA
15
on Algebra, Lemma 54.9. Since the complex is exact in (homological) degrees > m
we conclude that Hm is a finite R-module by More on Algebra, Lemma 54.3.
09PP
Lemma 6.10. Let R be a ring. Suppose that (A, d, γ) and (B, d, γ) are as in
Definition 6.4. Let ϕ : H0 (A) → H0 (B) be an R-algebra map. Assume
(1) A is a graded divided power polynomial algebra over R with finitely many
variables in each degree,
(2) Hk (B) = 0 for k > 0.
Then there exists a map ϕ : A → B of differential graded R-algebras compatible
with divided powers lifting ϕ.
Proof. Since A is obtained from R by adjoining divided power variables, we obtain
filtrations R ⊂ A(0) ⊂ A(1) ⊂ . . . such that A(m + 1) is obtained from A(m) by
adjoining finitely many divided power variables of degree m + 1. Then A(0) → S
is a surjection from a (usual) polynomial algebra over R onto S. Thus we can lift
ϕ to an R-algebra map ϕ(0) : A(0) → B(0).
Write A(1) = A(0)hT1 , . . . , Tm i for some divided power variables Tj of degree 1.
Let fj ∈ B0 be fj = ϕ(0)(d(Tj )). Observe that fj maps to zero in H0 (B) as dTj
maps to zero in H0 (A). Thus we can find bj ∈ B1 with d(bj ) = fj . By the universal
property of divided power polynomial algebras we find a lift ϕ(1) : A(1) → B of
ϕ(0) mapping Tj to fj .
Having constructed ϕ(m) for some m ≥ 1 we can construct ϕ(m+1) : A(m+1) → B
in exactly the same manner. We omit the details.
09PQ
Lemma 6.11. Let R be a Noetherian ring. Let R → S and R → T be finite type
ring maps. There exists a canonical structure of a divided power graded R-algebra
on
TorR
∗ (S, T )
Proof. Choose a factorization R → A → S as above. Since A → S is a quasiisomorphism and since Ad is a free R-module, we see that the differential graded
algebra B = A ⊗R T computes the tor groups displayed in the lemma. Choose
a surjection R[y1 , . . . , yk ] → T . Then we see that B is a quotient of the differential graded algebra A[y1 , . . . , yk ] whose homology sits in degree 0 (it is equal to
S[y1 , . . . , yk ]). By Lemma 6.6 the differential graded algebras B and A[y1 , . . . , yk ]
have divided power structures compatible with the differentials. Hence we obtain
our divided power structure on H(B) by Lemma 6.5.
The divided power algebra structure constructed in this way is independent of the
choice of A. Namely, if A0 is a second choice, then Lemma 6.10 implies there is
a map A → A0 preserving all structure and the augmentations towards S. Then
the induced map B = A ⊗R T → A0 ⊗R T 0 = B 0 is likewise and is a quasiisomorphism. The induced isomorphism of Tor algebras is therefore compatible
with all multiplication and divided powers.
7. Application to complete intersections
09PR
Let R be a ring. Let (A, d, γ) be as in Definition 6.4. A derivation of degree 2 is
an R-linear map θ : A → A with the following properties
(1) θ(Ad ) ⊂ Ad−2 ,
(2) θ(xy) = θ(x)y + xθ(y),
DIVIDED POWER ALGEBRA
16
(3) θ commutes with d,
(4) θ(γn (x)) = θ(x)γn−1 (x) for all x ∈ A2d all d.
In the following lemma we construct a derivation.
09PS
Lemma 7.1. Let R be a ring. Let (A, d, γ) be as in Definition 6.4. Let R0 → R be
a surjection of rings whose kernel has square zero and is generated by one element
f . If A is a graded divided power polynomial algebra over R with finitely many
variables in each degree, then we obtain a derivation θ : A/IA → A/IA where I is
the annihilator of f in R.
Proof. Since A is a divided power polynomial algebra, we can find a divided power
polynomial algebra A0 over R0 such that A = A0 ⊗R R0 . Moreover, we can lift d to
an R-linear operator d on A0 such that
(1) d(xy) = d(x)y + (−1)deg(x) xd(y) for x, y ∈ A0 homogeneous, and
(2) d(γn (x)) = d(x)γn−1 (x) for x ∈ A0even,+ .
We omit the details (hint: proceed one variable at the time). However, it may not
be the case that d2 is zero on A0 . It is clear that d2 maps A0 into f A0 ∼
= A/IA.
Hence d2 annihilates f A0 and factors as a map A → A/IA. Since d2 is R-linear
we obtain our map θ : A/IA → A/IA. The verification of the properties of a
derivation is immediate.
09PT
Lemma 7.2. Assumption and notation as in Lemma 7.1. Suppose S = H0 (A)
is isomorphic to R[x1 , . . . , xn ]/(f1 , . . . , fm ) for some n, m, and fj ∈ R[x1 , . . . , xn ].
Moreover, suppose given a relation
X
rj fj = 0
P
with rj ∈ R[x1 , . . . , xn ]. Choose rj0 , fj0 ∈ R0 [x1 , . . . , xn ] lifting rj , fj . Write rj0 fj0 =
gf for some g ∈ R/I[x1 , . . . , xn ]. If H1 (A) = 0 and all the coefficients of each rj
are in I, then there exists an element ξ ∈ H2 (A/IA) such that θ(ξ) = g in S/IS.
Proof. Let A(0) ⊂ A(1) ⊂ A(2) ⊂ . . . be the filtration of A such that A(m) is
gotten from A(m − 1) by adjoining divided power variables of degree m. Then A(0)
is a polynomial algebra over R equipped with an R-algebra surjection A(0) → S.
Thus we can choose a map
ϕ : R[x1 , . . . , xn ] → A(0)
lifting the augmentations to S. Next, A(1) = A(0)hT1 , . . . , Tt i forPsome divided
power variables Ti of degree 1. Since H0 (A) = S we can pick ξj ∈
A(0)Ti with
d(ξj ) = ϕ(fj ). Then
X
X
X
d
ϕ(rj )ξj =
ϕ(rj )ϕ(fj ) =
ϕ(rj fj ) = 0
P
Since H1 (A) = 0 we can pick ξ ∈ A2 with d(ξ) =
ϕ(rj )ξj . If the coefficients of
rj are in I, then the same is true for ϕ(rj ). In this case d(ξ) dies in A1 /IA1 and
hence ξ defines a class in H2 (A/IA).
The construction of θ in the proof of Lemma 7.1 proceeds by successively lifting
A(i) to A0 (i) and lifting the differential d. We lift ϕ to ϕ0 : R0 [x1 , . . . , xnP
] → A0 (0).
0
0
0
Next, we have A (1) = A (0)hT1 , . . . , Tt i. Moreover, we can lift ξj to ξj ∈
A0 (0)Ti .
0
0 0
0
0
0
Then d(ξj ) = ϕ (fj ) + f aj for some aj ∈ A (0). Consider a lift ξ ∈ A2 of ξ. Then
we know that
X
X
d(ξ 0 ) =
ϕ0 (rj0 )ξj0 +
f bi Ti
DIVIDED POWER ALGEBRA
17
for some bi ∈ A(0). Applying d again we find
X
X
X
θ(ξ) =
ϕ0 (rj0 )ϕ0 (fj0 ) +
f ϕ0 (rj0 )aj +
f bi d(Ti )
The first term gives us what we want. The second term is zero because the coefficients of rj are in I and hence are annihilated by f . The third term maps to zero
in H0 because d(Ti ) maps to zero.
The method of proof of the following lemma is apparantly due to Gulliksen.
09PU
Lemma 7.3. Let R0 → R be a surjection of Noetherian rings whose kernel has
square
P zero and is generated by one element f . Let S = R[x1 , . . . , xn ]/(f1 , . . . , fm ).
Let
rj fj = 0 be a relation in R[x1 , . . . , xm ]. Assume that
(1) each rj has coeffients in the annihilator I of fP
in R,
(2) for some lifts rj0 , fj0 ∈ R0 [x1 , . . . , xn ] we have
rj0 fj0 = gf where g is not
nilpotent in S.
Then S does not have finite tor dimension over R (i.e., S is not a perfect R-algebra).
Proof. Choose a Tate resolution R → A → S as in Lemma 6.8. Let ξ ∈ H2 (A/IA)
and θ : A/IA → A/IA be the element and derivation found in Lemmas 7.1 and
7.2. Observe that
θn (γn (ξ)) = g n
Hence if g is not nilpotent, then ξ n is nonzero in H2n (A/IA) for all n > 0. Since
H2n (A/IA) = TorR
2n (S, R/I) we conclude.
The following result can be found in [Rod88].
09PV
Lemma 7.4. Let (A, m) be a Noetherian local ring. Let I ⊂ J ⊂ A be proper
ideals. If A/J has finite tor dimension over A/I, then I/mI → J/mJ is injective.
Proof. Let f ∈ I be an element mapping to a nonzero element of I/mI which is
mapped to zero in J/mJ. We can choose an ideal I 0 with mI ⊂ I 0 ⊂ I such that I/I 0
is generated by the image ofPf . Set R = A/I and R0 = A/I 0 . Let J = (a1 , . . . , am )
for some aj ∈ A. Then f =
bj aj for some bj ∈ m. Let rj , fj ∈ R resp. rj0 , fj0 ∈ R0
be the image of bj , aj . Then we see we are in the situation of Lemma 7.3 (with the
ideal I of that lemma equal to mR ) and the lemma is proved.
09PW
Lemma 7.5. Let (A, m) be a Noetherian local ring. Let I ⊂ J ⊂ A be proper
ideals. Assume
(1) A/J has finite tor dimension over A/I, and
(2) J is generated by a regular sequence.
Then I is generated by a regular sequence and J/I is generated by a regular sequence.
Proof. By Lemma 7.4 we see that I/mI → J/mJ is injective. Thus we can find
s ≤ r and a minimal system of generators f1 , . . . , fr of J such that f1 , . . . , fs are in
I and form a minimal system of generators of I. The lemma follows as any minimal
system of generators of J is a regular sequence by More on Algebra, Lemmas 23.14
and 23.6.
09PX
Lemma 7.6. Let R → S be a local ring map of Noetherian local rings. Let I ⊂ R
and J ⊂ S be ideals with IS ⊂ J. If R → S is flat and S/mR S is regular, then the
following are equivalent
DIVIDED POWER ALGEBRA
18
(1) J is generated by a regular sequence and S/J has finite tor dimension as a
module over R/I,
(2) J is generated by a regular sequence and TorR/I
(S/J, R/mR ) is nonzero for
p
only finitely many p,
(3) I is generated by a regular sequence and J/IS is generated by a regular
sequence in S/IS.
Proof. If (3) holds, then J is generated by a regular sequence, see for example
More on Algebra, Lemmas 23.12 and 23.6. Moreover, if (3) holds, then S/J =
(S/I)/(J/I) has finite projective dimension over S/IS because the Koszul complex
will be a finite free resolution of S/J over S/IS. Since R/I → S/IS is flat, it then
follows that S/J has finite tor dimension over R/I by More on Algebra, Lemma
55.11. Thus (3) implies (1).
The implication (1) ⇒ (2) is trivial. Assume (2). By More on Algebra, Lemma
63.9 we find that S/J has finite tor dimension over S/IS. Thus we can apply
Lemma 7.5 to conclude that IS and J/IS are generated by regular sequences. Let
f1 , . . . , fr ∈ I be a minimal system of generators of I. Since R → S is flat, we see
that f1 , . . . , fr form a minimal system of generators for IS in S. Thus f1 , . . . , fr ∈ R
is a sequence of elements whose images in S form a regular sequence by More on
Algebra, Lemmas 23.14 and 23.6. Thus f1 , . . . , fr is a regular sequence in R by
Algebra, Lemma 67.5.
8. Local complete intersection rings
09PY
Let (A, m) be a Noetherian complete local ring. By the Cohen structure theorem
(see Algebra, Theorem 154.8) we can write A as the quotient of a regular Noetherian
complete local ring R. Let us say that A is a complete intersection if there exists
some surjection R → A with R a regular local ring such that the kernel is generated
by a regular sequence. The following lemma shows this notion is independent of
the choice of the surjection.
09PZ
Lemma 8.1. Let (A, m) be a Noetherian complete local ring. The following are
equivalent
(1) for every surjection of local rings R → A with R a regular local ring, the
kernel of R → A is generated by a regular sequence, and
(2) for some surjection of local rings R → A with R a regular local ring, the
kernel of R → A is generated by a regular sequence.
Proof. Let k be the residue field of A. If the characteristic of k is p > 0, then we
denote Λ a Cohen ring (Algebra, Definition 154.5) with residue field k (Algebra,
Lemma 154.6). If the characteristic of k is 0 we set Λ = k. Recall that Λ[[x1 , . . . , xn ]]
for any n is formally smooth over Z, resp. Q in the m-adic topology, see More
on Algebra, Lemma 30.1. Fix a surjection Λ[[x1 , . . . , xn ]] → A as in the Cohen
structure theorem (Algebra, Theorem 154.8).
Let R → A be a surjection from a regular local ring R. Let f1 , . . . , fr be a minimal
sequence of generators of Ker(R → A). We will use without further mention that
an ideal in a Noetherian local ring is generated by a regular sequence if and only
if any minimal set of generators is a regular sequence. Observe that f1 , . . . , fr is a
regular sequence in R if and only if f1 , . . . , fr is a regular sequence in the completion
DIVIDED POWER ALGEBRA
19
R∧ by Algebra, Lemmas 67.5 and 96.2. Moreover, we have
R∧ /(f1 , . . . , fr )R∧ = (R/(f1 , . . . , fn ))∧ = A∧ = A
because A is mA -adically complete (first equality by Algebra, Lemma 96.1). Finally,
the ring R∧ is regular since R is regular (More on Algebra, Lemma 34.4). Hence
we may assume R is complete.
If R is complete we can choose a map Λ[[x1 , . . . , xn ]] → R lifting the given map
Λ[[x1 , . . . , xn ]] → A, see More on Algebra, Lemma 29.5. By adding some more
variables y1 , . . . , ym mapping to generators of the kernel of R → A we may assume
that Λ[[x1 , . . . , xn , y1 , . . . , ym ]] → R is surjective (some details omitted). Then we
can consider the commutative diagram
Λ[[x1 , . . . , xn , y1 , . . . , ym ]]
/R
Λ[[x1 , . . . , xn ]]
/A
By Algebra, Lemma 133.6 we see that the condition for R → A is equivalent to the
condition for the fixed chosen map Λ[[x1 , . . . , xn ]] → A. This finishes the proof of
the lemma.
The following two lemmas are sanity checks on the definition given above.
09Q0
Lemma 8.2. Let R be a regular ring. Let p ⊂ R be a prime. Let f1 , . . . , fr ∈ p be
a regular sequence. Then the completion of
A = (R/(f1 , . . . , fr ))p = Rp /(f1 , . . . , fr )Rp
is a complete intersection in the sense defined above.
Proof. The completion of A is equal to A∧ = Rp∧ /(f1 , . . . , fr )Rp∧ because completion for finite modules over the Noetherian ring Rp is exact (Algebra, Lemma
96.1). The image of the sequence f1 , . . . , fr in Rp is a regular sequence by Algebra,
Lemmas 96.2 and 67.5. Moreover, Rp∧ is a regular local ring by More on Algebra,
Lemma 34.4. Hence the result holds by our definition of complete intersection for
complete local rings.
The following lemma is the analogue of Algebra, Lemma 133.4.
09Q1
Lemma 8.3. Let R be a regular ring. Let p ⊂ R be a prime. Let I ⊂ p be an
ideal. Set A = (R/I)p = Rp /Ip . The following are equivalent
(1)
(2)
(3)
(4)
the completion of A is a complete intersection in the sense above,
Ip ⊂ Rp is generated by a regular sequence,
the module (I/I 2 )p can be generated by dim(Rp ) − dim(A) elements,
add more here.
Proof. We may and do replace R by its localization at p. Then p = m is the
maximal ideal of R and A = R/I. Let f1 , . . . , fr ∈ I be a minimal sequence
of generators. The completion of A is equal to A∧ = R∧ /(f1 , . . . , fr )R∧ because
completion for finite modules over the Noetherian ring Rp is exact (Algebra, Lemma
96.1).
DIVIDED POWER ALGEBRA
20
If (1) holds, then the image of the sequence f1 , . . . , fr in R∧ is a regular sequence
by assumption. Hence it is a regular sequence in R by Algebra, Lemmas 96.2 and
67.5. Thus (1) implies (2).
Assume (3) holds. Set c = dim(R)−dim(A) and let f1 , . . . , fc ∈ I map to generators
of I/I 2 . by Nakayama’s lemme (Algebra, Lemma 19.1) we see that I = (f1 , . . . , fc ).
Since R is regular and hence Cohen-Macaulay (Algebra, Proposition 102.6) we see
that f1 , . . . , fc is a regular sequence by Algebra, Proposition 102.6. Thus (3) implies
(2). Finally, (2) implies (1) by Lemma 8.2.
The following result is due to Avramov, see [Avr75].
09Q2
Proposition 8.4. Let A → B be a flat local homomorphism of Noetherian local
rings. Then the following are equivalent
(1) B ∧ is a complete intersection,
(2) A∧ and (B/mA B)∧ are complete intersections.
Proof. Consider the diagram
BO
/ B∧
O
/ A∧
A
Since the horizontal maps are faithfully flat (Algebra, Lemma 96.3) we conclude that
the right vertical arrow is flat (for example by Algebra, Lemma 98.15). Moreover,
we have (B/mA B)∧ = B ∧ /mA∧ B ∧ by Algebra, Lemma 96.1. Thus we may assume
A and B are complete local Noetherian rings.
Assume A and B are complete local Noetherian rings. Choose a diagram
SO
/B
O
R
/A
as in More on Algebra, Lemma 30.3. Let I = Ker(R → A) and J = Ker(S → B).
Note that since R/I = A → B = S/J is flat the map J/I ⊗R R/mR → J/J ∩mR S is
an isomorphism. Hence a minimal system of generators of J/I maps to a minimal
system of generators of Ker(S/mR S → B/mA B). Finally, S/mR S is a regular local
ring.
Assume (1) holds, i.e., J is generated by a regular sequence. Since A = R/I → B =
S/J is flat we see Lemma 7.6 applies and we deduce that I and J/I are generated
by regular sequences. We have dim(B) = dim(A)+dim(B/mA B) and dim(S/IS) =
dim(A) + dim(S/mR S) (Algebra, Lemma 111.7). Thus J/I is generated by
dim(S/J) − dim(S/IS) = dim(S/mR S) − dim(B/mA B)
elements (Algebra, Lemma 59.12). It follows that Ker(S/mR S → B/mA B) is generated by the same number of elements (see above). Hence Ker(S/mR S → B/mA B)
is generated by a regular sequence, see for example Lemma 8.3. In this way we see
that (2) holds.
If (2) holds, then I and J/J ∩mR S are generated by regular sequences. Lifting these
generators (see above), using flatness of R/I → S/IS, and using Grothendieck’s
DIVIDED POWER ALGEBRA
21
lemma (Algebra, Lemma 98.3) we find that J/I is generated by a regular sequence
in S/IS. Thus Lemma 7.6 tells us that J is generated by a regular sequence, whence
(1) holds.
09Q3
Definition 8.5. Let A be a Noetherian ring.
(1) If A is local, then we say A is a complete intersection if its completion is a
complete intersection in the sense above.
(2) In general we say A is a local complete intersection if all of its local rings
are complete intersections.
We will check below that this does not conflict with the terminology introduced
in Algebra, Definitions 133.1 and 133.5. But first, we show this “makes sense” by
showing that if A is a Noetherian local complete intersection, then A is a local
complete intersection, i.e., all of its local rings are complete intersections.
09Q4
Lemma 8.6. Let (A, m) be a Noetherian local ring. Let p ⊂ A be a prime ideal.
If A is a complete intersection, then Ap is a complete intersection too.
Proof. Choose a prime q of A∧ lying over p (this is possible as A → A∧ is faithfully
flat by Algebra, Lemma 96.3). Then Ap → (A∧ )q is a flat local ring homomorphism.
Thus by Proposition 8.4 we see that Ap is a complete intersection if and only if
(A∧ )q is a complete intersection. Thus it suffices to prove the lemma in case A is
complete (this is the key step of the proof).
Assume A is complete. By definition we may write A = R/(f1 , . . . , fr ) for some
regular sequence f1 , . . . , fr in a regular local ring R. Let q ⊂ R be the prime
corresponding to p. Observe that f1 , . . . , fr ∈ q and that Ap = Rq /(f1 , . . . , fr )Rq .
Hence Ap is a complete intersection by Lemma 8.2.
09Q5
Lemma 8.7. Let A be a Noetherian ring. Then A is a local complete intersection
if and only if Am is a complete intersection for every maximal ideal m of A.
Proof. This follows immediately from Lemma 8.6 and the definitions.
09Q6
Lemma 8.8. Let S be a finite type algebra over a field k.
(1) for a prime q ⊂ S the local ring Sq is a complete intersection in the sense
of Algebra, Definition 133.5 if and only if Sq is a complete intersection in
the sense of Definition 8.5, and
(2) S is a local complete intersection in the sense of Algebra, Definition 133.1
if and only if S is a local complete intersection in the sense of Definition
8.5.
Proof. Proof of (1). Let k[x1 , . . . , xn ] → S be a surjection. Let p ⊂ k[x1 , . . . , xn ]
be the prime ideal corresponding to q. Let I ⊂ k[x1 , . . . , xn ] be the kernel of our
surjection. Note that k[x1 , . . . , xn ]p → Sq is surjective with kernel Ip . Observe that
k[x1 , . . . , xn ] is a regular ring by Algebra, Proposition 113.2. Hence the equivalence
of the two notions in (1) follows by combining Lemma 8.3 with Algebra, Lemma
133.7.
Having proved (1) the equivalence in (2) follows from the definition and Algebra,
Lemma 133.9.
09Q7
Lemma 8.9. Let A → B be a flat local homomorphism of Noetherian local rings.
Then the following are equivalent
DIVIDED POWER ALGEBRA
22
(1) B is a complete intersection,
(2) A and B/mA B are complete intersections.
Proof. Now that the definition makes sense this is a trivial reformulation of the
(nontrivial) Proposition 8.4.
9. Local complete intersection maps
09Q9
Let A → B be a local homomorphism of Noetherian complete local rings. A
consequence of the Cohen structure theorem is that we can find a commutative
diagram
/B
S_
O
A
of Noetherian complete local rings with S → B surjective, A → S flat, and S/mA S
a regular local ring. This follows from More on Algebra, Lemma 30.3. Let us
(temporarily) say A → S → B is a good factorization of A → B if S is a Noetherian
local ring, A → S → B are local ring maps, S → B surjective, A → S flat, and
S/mA S regular. Let us say that A → B is a complete intersection homomorphism
if there exists some good factorization A → S → B such that the kernel of S → B
is generated by a regular sequence. The following lemma shows this notion is
independent of the choice of the diagram.
09QA
Lemma 9.1. Let A → B be a local homomorphism of Noetherian complete local
rings. The following are equivalent
(1) for some good factorization A → S → B the kernel of S → B is generated
by a regular sequence, and
(2) for every good factorization A → S → B the kernel of S → B is generated
by a regular sequence.
Proof. Let A → S → B be a good factorization. As B is complete we obtain a
factorization A → S ∧ → B where S ∧ is the completion of S. Note that this is
also a good factorization: The ring map S → S ∧ is flat (Algebra, Lemma 96.2),
hence A → S ∧ is flat. The ring S ∧ /mA S ∧ = (S/mA S)∧ is regular since S/mA S is
regular (More on Algebra, Lemma 34.4). Let f1 , . . . , fr be a minimal sequence of
generators of Ker(S → B). We will use without further mention that an ideal in a
Noetherian local ring is generated by a regular sequence if and only if any minimal
set of generators is a regular sequence. Observe that f1 , . . . , fr is a regular sequence
in S if and only if f1 , . . . , fr is a regular sequence in the completion S ∧ by Algebra,
Lemma 67.5. Moreover, we have
S ∧ /(f1 , . . . , fr )R∧ = (S/(f1 , . . . , fn ))∧ = B ∧ = B
because B is mB -adically complete (first equality by Algebra, Lemma 96.1). Thus
the kernel of S → B is generated by a regular sequence if and only if the kernel
of S ∧ → B is generated by a regular sequence. Hence it suffices to consider good
factorizations where S is complete.
Assume we have two factorizations A → S → B and A → S 0 → B with S and
S 0 complete. By More on Algebra, Lemma 30.4 the ring S ×B S 0 is a Noetherian
complete local ring. Hence, using More on Algebra, Lemma 30.3 we can choose
DIVIDED POWER ALGEBRA
23
a good factorization A → S 00 → S ×B S 0 with S 00 complete. Thus it suffices to
show: If A → S 0 → S → B are comparable good factorizations, then Ker(S → B)
is generated by a regular sequence if and only if Ker(S 0 → B) is generated by a
regular sequence.
Let A → S 0 → S → B be comparable good factorizations. First, since S 0 /mR S 0 →
S/mR S is a surjection of regular local rings, the kernel is generated by a regular
sequence x1 , . . . , xc ∈ mS 0 /mR S 0 which can be extended to a regular system of
parameters for the regular local ring S 0 /mR S 0 , see (Algebra, Lemma 105.4). Set
I = Ker(S 0 → S). By flatness of S over R we have
I/mR I = Ker(S 0 /mR S 0 → S/mR S) = (x1 , . . . , xc ).
Choose lifts x1 , . . . , xc ∈ I. These lifts form a regular sequence generating I as S 0
is flat over R, see Algebra, Lemma 98.3.
We conclude that if also Ker(S → B) is generated by a regular sequence, then so
is Ker(S 0 → B), see More on Algebra, Lemmas 23.12 and 23.6.
Conversely, assume that J = Ker(S 0 → B) is generated by a regular sequence.
Because the generators x1 , . . . , xc of I map to linearly independent elements of
mS 0 /m2S 0 we see that I/mS 0 I → J/mS 0 J is injective. Hence there exists a minimal
system of generators x1 , . . . , xc , y1 , . . . , yd for J. Then x1 , . . . , xc , y1 , . . . , yd is a
regular sequence and it follows that the images of y1 , . . . , yd in S form a regular
sequence generating Ker(S → B). This finishes the proof of the lemma.
In the following proposition observe that the condition on vanishing of Tor’s applies
in particular if B has finite tor dimension over A and thus in particular if B is flat
over A.
09QB
Proposition 9.2. Let A → B be a local homomorphism of Noetherian local rings.
Then the following are equivalent
(1) B is a complete intersection and TorA
p (B, A/mA ) is nonzero for only finitely
many p,
(2) A is a complete intersection and A∧ → B ∧ is a complete intersection homomorphism in the sense defined above.
Proof. Let F• → A/mA be a resolution by finite free A-modules. Observe that
∧
∧
TorA
p (B, A/mA ) is the pth homology of the complex F• ⊗A B. Let F• = F• ⊗A A
∧
∧
∧
be the completion. Then F• is a resolution of A /mA∧ by finite free A -modules
(as A → A∧ is flat and completion on finite modules is exact, see Algebra, Lemmas
96.1 and 96.2). It follows that
F•∧ ⊗A∧ B ∧ = F• ⊗A B ⊗B B ∧
By flatness of B → B ∧ we conclude that
∧
A
∧
∧
∧
TorA
p (B , A /mA∧ ) = Torp (B, A/mA ) ⊗B B
In this way we see that the condition in (1) on the local ring map A → B is
equivalent to the same condition for the local ring map A∧ → B ∧ . Thus we may
assume A and B are complete local Noetherian rings (since the other conditions
are formulated in terms of the completions in any case).
DIVIDED POWER ALGEBRA
24
Assume A and B are complete local Noetherian rings. Choose a diagram
SO
/B
O
R
/A
as in More on Algebra, Lemma 30.3. Let I = Ker(R → A) and J = Ker(S → B).
The proposition now follows from Lemma 7.6.
09QC
Remark 9.3. It appears difficult to define an good notion of “local complete intersection homomorphisms” for maps between general Noetherian rings. The reason
is that, for a local Noetherian ring A, the fibres of A → A∧ are not local complete
intersection rings. Thus, if A → B is a local homomorphism of local Noetherian
rings, and the map of completions A∧ → B ∧ is a complete intersection homomorphism in the sense defined above, then (Ap )∧ → (Bq )∧ is in general not a complete
intersection homomorphism in the sense defined above. A solution can be had
by working exclusively with excellent Noetherian rings. More generally, one could
work with those Noetherian rings whose formal fibres are complete intersections,
see [Rod87]. We will develop this theory in Dualizing Complexes, Section 42.
To finish of this section we compare the notion defined above with the notion
introduced in More on Algebra, Section 8.
09QD
Lemma 9.4. Consider a commutative diagram
S_
/B
O
A
of Noetherian local rings with S → B surjective, A → S flat, and S/mA S a regular
local ring. The following are equivalent
(1) Ker(S → B) is generated by a regular sequence, and
(2) A∧ → B ∧ is a complete intersection homomorphism as defined above.
Proof. Omitted. Hint: the proof is indentical to the argument given in the first
paragraph of the proof of Lemma 9.1.
09QE
Lemma 9.5. Let A be a Noetherian ring. Let A → B be a finite type ring map.
The following are equivalent
(1) A → B is a local complete intersection in the sense of More on Algebra,
Definition 25.2,
(2) for every prime q ⊂ B and with p = A ∩ q the ring map (Ap )∧ → (Bq )∧ is
a complete intersection homomorphism in the sense defined above.
Proof. Choose a surjection R = A[x1 , . . . , xn ] → B. Observe that A → R is flat
with regular fibres. Let I be the kernel of R → B. Assume (2). Then we see that
I is locally generated by a regular sequence by Lemma 9.4 and Algebra, Lemma
67.6. In other words, (1) holds. Conversely, assume (1). Then after localizing on R
and B we can assume that I is generated by a Koszul regular sequence. By More
on Algebra, Lemma 23.6 we find that I is locally generated by a regular sequence.
Hence (2) hold by Lemma 9.4. Some details omitted.
DIVIDED POWER ALGEBRA
09QF
25
Lemma 9.6. Let A be a Noetherian ring. Let A → B be a finite type ring map
such that the image of Spec(B) → Spec(A) contains all closed points of Spec(A).
Then the following are equivalent
(1) B is a complete intersection and A → B has finite tor dimension,
(2) A is a complete intersection and A → B is a complete intersection in the
sense of More on Algebra, Definition 25.2.
Proof. This is a reformulation of Proposition 9.2 via Lemma 9.5. We omit the
details.
10. Other chapters
Preliminaries
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
(16)
(17)
(18)
(19)
(20)
(21)
(22)
(23)
(24)
Introduction
Conventions
Set Theory
Categories
Topology
Sheaves on Spaces
Sites and Sheaves
Stacks
Fields
Commutative Algebra
Brauer Groups
Homological Algebra
Derived Categories
Simplicial Methods
More on Algebra
Smoothing Ring Maps
Sheaves of Modules
Modules on Sites
Injectives
Cohomology of Sheaves
Cohomology on Sites
Differential Graded Algebra
Divided Power Algebra
Hypercoverings
Schemes
(25)
(26)
(27)
(28)
(29)
(30)
(31)
(32)
(33)
(34)
Schemes
Constructions of Schemes
Properties of Schemes
Morphisms of Schemes
Cohomology of Schemes
Divisors
Limits of Schemes
Varieties
Topologies on Schemes
Descent
(35)
(36)
(37)
(38)
(39)
(40)
Derived Categories of Schemes
More on Morphisms
More on Flatness
Groupoid Schemes
More on Groupoid Schemes
Étale Morphisms of Schemes
Topics in Scheme Theory
(41)
(42)
(43)
(44)
(45)
(46)
(47)
(48)
(49)
(50)
(51)
(52)
Chow Homology
Intersection Theory
Picard Schemes of Curves
Adequate Modules
Dualizing Complexes
Algebraic Curves
Resolution of Surfaces
Semistable Reduction
Fundamental Groups of Schemes
Étale Cohomology
Crystalline Cohomology
Pro-étale Cohomology
Algebraic Spaces
(53)
(54)
(55)
(56)
(57)
(58)
(59)
(60)
(61)
(62)
(63)
(64)
(65)
(66)
(67)
(68)
Algebraic Spaces
Properties of Algebraic Spaces
Morphisms of Algebraic Spaces
Decent Algebraic Spaces
Cohomology of Algebraic Spaces
Limits of Algebraic Spaces
Divisors on Algebraic Spaces
Algebraic Spaces over Fields
Topologies on Algebraic Spaces
Descent and Algebraic Spaces
Derived Categories of Spaces
More on Morphisms of Spaces
Pushouts of Algebraic Spaces
Groupoids in Algebraic Spaces
More on Groupoids in Spaces
Bootstrap
DIVIDED POWER ALGEBRA
Topics in Geometry
(69)
(70)
(71)
(72)
(73)
Quotients of Groupoids
Simplicial Spaces
Formal Algebraic Spaces
Restricted Power Series
Resolution of Surfaces Revisited
Deformation Theory
(74) Formal Deformation Theory
(75) Deformation Theory
(76) The Cotangent Complex
Algebraic Stacks
(77)
(78)
(79)
(80)
(81)
Algebraic Stacks
Examples of Stacks
Sheaves on Algebraic Stacks
Criteria for Representability
Artin’s Axioms
26
(82) Quot and Hilbert Spaces
(83) Properties of Algebraic Stacks
(84) Morphisms of Algebraic Stacks
(85) Cohomology of Algebraic Stacks
(86) Derived Categories of Stacks
(87) Introducing Algebraic Stacks
(88) More on Morphisms of Stacks
Miscellany
(89) Examples
(90) Exercises
(91) Guide to Literature
(92) Desirables
(93) Coding Style
(94) Obsolete
(95) GNU Free Documentation License
(96) Auto Generated Index
References
[Avr75] Luchezar L. Avramov, Flat morphisms of complete intersections, Dokl. Akad. Nauk SSSR
225 (1975), no. 1, 11–14.
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